Chapter 15, Problem 34QAP

Solid lead nitrate is added to a solution that is 0.020 M in OH^- and SO₄^2-. Addition of the lead nitrate does not change the volume of the solution.

(a) Which compound, PbSO₄ or Pb(OH)₂ $\left(K_{\text{sp}}=2.8\times {10}^{-16}\right)$, will precipitate first?

(b) What is the pH of the solution when PbSO₄ first starts to precipitate?

Interpretation Introduction

Interpretation: Whether ${\text{PbSO}}_{\text{4}}\text{or}\text{Pb}{\left(\text{OH}\right)}_{\text{2}}\left({\text{K}}_{\text{sp}}\text{=2}{\text{.8×10}}^{\text{-16}}\right)\text{,}$ will precipitate first needs to be determined.

Concept Introduction :

${\text{PbSO}}_{\text{4}}$ (Equilibrium reaction)

${\text{PbSO}}_{\text{4}}\left(\text{s}\right)\rightleftharpoons {\text{Pb}}^{\text{+2}}\left(\text{aq}\right)+{\text{SO}}_{\text{4}}{}^{\text{2}-}\left(\text{aq}\right)$

Expression of ${\text{Pb}}^{\text{+2}}$ required for precipitation of ${\text{PbSO}}_{\text{4}}$ is −

$K_{sp}=\left[{\text{Pb}}^{\text{+2}}\right]\left[{\text{SO}}_{\text{4}}{}^{\text{2}-}\right]$...... (1)

Where, $K_{sp}$ is solubility product of ${\text{PbSO}}_{\text{4}}$ , $\left[{\text{Pb}}^{\text{+2}}\right]$ is concentration of lead ion and $\left[{\text{SO}}_{\text{4}}{}^{\text{2}-}\right]$ concentration of sulfate ion.

  $\text{Pb}{\left(\text{OH}\right)}_2$ (Equilibrium reaction)

  $\text{Pb}{\left(\text{OH}\right)}_2\left(\text{s}\right)\rightleftharpoons {\text{Pb}}^{\text{+2}}\left(\text{aq}\right)+2{\text{OH}}^{-}\left(\text{aq}\right)$

Expression of ${\text{Pb}}^{\text{+2}}$ required for precipitation of $\text{Pb}{\left(\text{OH}\right)}_2$ is −

$K_{sp}=\left[{\text{Pb}}^{\text{+2}}\right]{\left[{\text{OH}}^{-}\right]}^2$...... (2)

Where, $K_{sp}$ is solubility product of $\text{Pb}{\left(\text{OH}\right)}_2$ , $\left[{\text{Pb}}^{\text{+2}}\right]$ is concentration of lead ion and $\left[{\text{OH}}^{-}\right]$ concentration of hydroxide ion.

Answer to Problem 34QAP

The compound which precipitates first is $\text{Pb}{\left(\text{OH}\right)}_2$ .

Explanation of Solution

Given Information:

$\begin{array}{l}{\text{OH}}^{-}=\text{0}\text{.020}\text{M}\\ {\text{SO}}_{\text{4}}^{\text{2}-}=\text{0}\text{.020}\text{M}\end{array}$

The value of solubility product for both the salts is $2.8\times {10}^{-16}$ .

Calculation:

  ${\text{PbSO}}_{\text{4}}$ (Equilibrium reaction) is as follows:

${\text{PbSO}}_{\text{4}}\left(\text{s}\right)\rightleftharpoons {\text{Pb}}^{\text{+2}}\left(\text{aq}\right)+{\text{SO}}_{\text{4}}{}^{\text{2}-}\left(\text{aq}\right)$

Expression of ${\text{Pb}}^{\text{+2}}$ required for precipitation of ${\text{PbSO}}_{\text{4}}$ is −

$K_{sp}=\left[{\text{Pb}}^{\text{+2}}\right]\left[{\text{SO}}_{\text{4}}{}^{\text{2}-}\right]$...... (1)

Where, $K_{sp}$ is solubility product of ${\text{PbSO}}_{\text{4}}$ , $\left[{\text{Pb}}^{\text{+2}}\right]$ is concentration of lead ion and $\left[{\text{SO}}_{\text{4}}{}^{\text{2}-}\right]$ concentration of sulfate ion.

  $K_{sp}\left({\text{PbSO}}_4\right)=2.8\times {10}^{-16}$

${\text{SO}}_{\text{4}}^{\text{2}-}=\text{0}\text{.020}\text{M}$

  $\left[{\text{Pb}}^{\text{+2}}\right]$ is find out by putting above values in equation (1).

  $\begin{array}{c}{K}_{sp}=\left[{\text{Pb}}^{\text{+2}}\right]\left[{\text{SO}}_{\text{4}}{}^{\text{2}-}\right]\\ 2.8\times {10}^{-16}=\left[{\text{Pb}}^{\text{+2}}\right]\times 0.020\\ \left[{\text{Pb}}^{\text{+2}}\right]=\frac{2.8\times {10}^{-16}}{0.020}\\ \left[{\text{Pb}}^{\text{+2}}\right]=1.4\times {10}^{-14}\text{M}\end{array}$

So, $\left[{\text{Pb}}^{\text{+2}}\right]$ concentration is $1.4\times {10}^{-14}\text{M}$ .

  $\text{Pb}{\left(\text{OH}\right)}_2$ (Equilibrium reaction) is as follows:

$\text{Pb}{\left(\text{OH}\right)}_2\left(\text{s}\right)\rightleftharpoons {\text{Pb}}^{\text{+2}}\left(\text{aq}\right)+2{\text{OH}}^{-}\left(\text{aq}\right)$

Expression of

${\text{Pb}}^{\text{+2}}$ required for precipitation of $\text{Pb}{\left(\text{OH}\right)}_2$ is −

$K_{sp}=\left[{\text{Pb}}^{\text{+2}}\right]{\left[{\text{OH}}^{-}\right]}^2$...... (2)

Where, $K_{sp}$ is solubility product of $\text{Pb}{\left(\text{OH}\right)}_2$ , $\left[{\text{Pb}}^{\text{+2}}\right]$ is concentration of lead ion and $\left[{\text{OH}}^{-}\right]$ concentration of hydroxide ion.

  $K_{sp}\left(\text{Pb}{\left(\text{OH}\right)}_2\right)=2.8\times {10}^{-16}$

$\left[{\text{OH}}^{-}\right]=\text{0}\text{.020}\text{M}$

  $\left[{\text{Pb}}^{\text{+2}}\right]$ is find out by substitution of above values in equation (2).

  $\begin{array}{c}{K}_{sp}=\left[{\text{Pb}}^{\text{+2}}\right]{\left[{\text{OH}}^{-}\right]}^2\\ 2.8\times {10}^{-16}=\left[{\text{Pb}}^{\text{+2}}\right]\times {\left(0.020\right)}^2\\ \left[{\text{Pb}}^{\text{+2}}\right]=\frac{2.8\times {10}^{-16}}{0.0004}\\ \left[{\text{Pb}}^{\text{+2}}\right]=7\times {10}^{-13}\text{M}\end{array}$

So, $\left[{\text{Pb}}^{\text{+2}}\right]$ concentration is $7\times {10}^{-13}\text{M}$ .

$\left[{\text{Pb}}^{\text{+2}}\right]$ Required for precipitation of ${\text{PbSO}}_{\text{4}}$ is $1.4\times {10}^{-14}\text{M}$ which is less than $\left[{\text{Pb}}^{\text{+2}}\right]$ required for precipitation of $\text{Pb}{\left(\text{OH}\right)}_2$ that is $7\times {10}^{-13}\text{M}$ .

Hence, $\text{Pb}{\left(\text{OH}\right)}_2$ precipitates first.

Interpretation Introduction

Interpretation:The pH of the solution when ${\text{PbSO}}_{\text{4}}$ first starts to precipitate needs to be determined.

Concept Introduction :

Formula to calculate $\text{pH}$ from $\text{pOH}$ value is as follows:

$\text{pH=14}-\text{pOH}$

Formula to calculate $\text{pOH}$ is as follows:

$\text{pOH=}-\log \left[{\text{OH}}^{-}\right]$

Answer to Problem 34QAP

When ${\text{PbSO}}_{\text{4}}$ first starts to precipitate pH of the solution is $\text{9}\text{.24}$ .

Explanation of Solution

When ${\text{PbSO}}_{\text{4}}$ first start to precipitate, value of $\left[{\text{OH}}^{-}\right]$ can be calculated as follows:

From $\text{Pb}{\left(\text{OH}\right)}_2$ dissociation reaction,

  $\begin{array}{c}{K}_{sp}=\left[{\text{Pb}}^{\text{+2}}\right]{\left[{\text{OH}}^{-}\right]}^2\\ \left[{\text{OH}}^{-}\right]={\left(\frac{K_{sp}}{\left[{\text{Pb}}^{\text{+2}}\right]}\right)}^{\frac{1}{2}}\\ \left[{\text{OH}}^{-}\right]={\left(\frac{2.8\times {10}^{-16}}{9\times {10}^{-7}}\right)}^{\frac{1}{2}}\\ =0.0000176\text{M}\end{array}$

So, $\left[{\text{OH}}^{-}\right]$ is $0.0000176\text{M}$

Calculation of $\text{pOH}$ .

$\begin{array}{c}\text{pOH=}-\log \left[{\text{OH}}^{-}\right]\\ =-\log \left(0.0000176\right)\\ =4.75\end{array}$

So, value of $\text{pOH}$ is $4.75$ .

Calculation of $\text{pOH}$ of solution as follows:

  $\begin{array}{c}\text{pH=14}-\text{pOH}\\ \text{=14}-4.75\\ =9.24\end{array}$

Hence, $\text{pH}$ of the solution is $9.24$ .